Note that, if every projective module is simple. Then, in particular, $_AA$ is simple. Thus, $A$ is simple ring. (I am assuming that $A$ is finite dimensional $K$-algebra with identity)
However, if every indecomposable projective left $A$-module is simple, then by using Krull-Schmidt theorem($_AA$ is a direct sum of indecomposable projective modules and these indecomposable summands are uniquely determined up to isomorphism and permutation), $A$ is a direct sum of indecomposable projective left $A$- modules. Hence, direct sum of simple modules. Thus, $A$ is semisimple.
Following the discussion in the comments and considering the title of your question. If every simple module is projective, then by Krull Schmidt theorem and the fact that indecomposable projectives have simple head. We obtain that the simples are isomorphic to their projective cover. Thus, $A$ is a direct sum of simples.